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Question

If In=π/40tann(x)dx then I2+I4,+I3+I5,+I4+I6,... are in ?

A
A.P
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B
G.P
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C
H.P
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D
A.G.P
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Solution

The correct option is D H.P
In=π/40tann(x)dx
In+In+2=π/40(tann(x)+tann+2(x))dx=π/40tann(x)(1+tan2x)dx=π/40tann(x)sec2xdx
Let tanx=t
sec2xdx=dt
π/40tann(x)sec2xdx=10tndt
10tndt=1n+1
So, the terms are 1/3,1/4,1/5...(H.P)

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